Let $\Phi$ be a (crystallographic) root system with Weyl group $\mathcal{W}$, and $\Phi^+$ a choice of positive roots, and $$ q := \prod_{\alpha\in\Phi^+} (\exp(\alpha/2) - \exp(-\alpha/2)) = \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w\rho) $$ be the denominator in the Weyl character formula ("WCF"); here, of course, $\exp$ is a formal exponential, $\rho := \frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha = \sum_i\varpi_i$ is the Weyl vector, $\varpi_i$ are the fundamental weights, and $\mathrm{sgn}$ is the abelian character of $\mathcal{W}$ with value $-1$ on the reflections.

Since $q$ is $\mathcal{W}$-anti-invariant (we have $w(q) = \mathrm{sgn}(w)\,q$), it follows that $q^2$ is $\mathcal{W}$-invariant. So $q^2$ can be expressed as a polynomial in the fundamental characters (the fundamental characters being the $x_i := q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ by the WCF); or equivalently, as a polynomial in the averages of the fundamental weights (meaning the $\frac{1}{\#\mathcal{W}} \sum_{w\in\mathcal{W}} \exp(w\varpi_i)$): see the references to Bourbaki and Lorenz in this related question. This invariant quantity $q^2$ can be considered as a kind of discriminant (see PS3 below).

My question about this square denominator $q^2$ is: how can we compute it in practice (as a polynomial of the kind I just described)? Is there a convenient expression? Or perhaps, can I get LiE or Sage to compute it?

I am also interested in any sort of remarks about it: for example, does it have a standard name (beyond "the square of the Weyl denominator")? Is it the (virtual) character of some naturally defined element in the Grothendieck group of representations?

PS 1: I should have made it clear that I am looking for a computational approach that does not involve writing down all the $\#\mathcal{W}$ terms in $q$. (LiE is capable of doing computations on the ring of [multiplicative] $\mathcal{W}$-invariants without fully expanding them; so the question is whether we can do this for a $\mathcal{W}$-anti-invariant like $q$.)

PS 2: If we let $R = \mathbb{C}[\exp(\Lambda)]^{\mathcal{W}}$, where $\Lambda := \langle\Phi^\vee\rangle^*$ is the weight lattice, be the ring of multiplicative $\mathcal{W}$-invariants (which is a polynomial ring), then the ring of multiplicative $\mathcal{W}_0$-invariants, where $\mathcal{W}_0 := \ker(\mathrm{sgn})$ is the group of rotations in the Weyl group, is the free quadratic algebra $R \oplus R q$. So asking to describe $q^2$ in $R$ is the missing bit in the presentation of this algebra. Hopefully this helps motivate the question.

PS 3: In the case where $\Phi = A_n$, then $q^2$ expressed as a polynomial of $x_1,\ldots,x_n$ is exactly the discriminant of the polynomial (in the $z$ variable) $z^{n+1} - x_1 z^n + x_2 z^{n-1} + \cdots + (-1)^n x_n z + (-1)^{n+1}$ (indeed, consider a diagonal element of $SL_{n+1}$: then $x_1,\ldots,x_n$ give the elementary symmetric functions of the $n+1$ eigenvalues, whose product is $1$, and $q^2$ is their discriminant since $q$ is the Vandermonde determinant). What I'm asking for is a generalization of this to the other root systems.

  • $\begingroup$ I don't understand "$q^2$ can be expressed as a polynomial of the fundamental characters ($q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$)". Are you saying that $q^2$ equals your expression with denominator $q$, so that $q^3$ equals $\sum_{w \in \mathcal W} \operatorname{sgn}(w)\exp(w(\rho + \varpi_i))$? $\endgroup$
    – LSpice
    Jun 1, 2017 at 22:43
  • 1
    $\begingroup$ @LSpice I'm saying that if we let $x_i:=q^{-1} \sum_w\mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ (the character with highest weight $\varpi_i$: this is just the Weyl character formula), then $q^2$ is some polynomial in $x_1,\ldots,x_n$ where $n$ is the rank of $\Phi$. This does have $q$ in the denominator, but it just means it cancels somehow (that's the whole point of the WCF). $\endgroup$
    – Gro-Tsen
    Jun 2, 2017 at 7:32
  • 1
    $\begingroup$ For example, for $A_2$, we have $q^2 = x_1^2 x_2^2 - 4 x_1^3 - 4 x_2^3 + 18 x_1 x_2 - 27$ (the standard cubic discriminant when the product of the three roots is $1$) where $x_i = q^{-1} \sum_w \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ and $\rho=\varpi_1+\varpi_2$. $\endgroup$
    – Gro-Tsen
    Jun 2, 2017 at 7:35
  • $\begingroup$ Oh, I see; the parenthetical is defining fundamental characters, not exhibiting $q^2$ as such a polynomial. $\endgroup$
    – LSpice
    Jun 2, 2017 at 14:02
  • $\begingroup$ @LSpice I agree that my wording was very unclear. I'll try to improve it somewhat. $\endgroup$
    – Gro-Tsen
    Jun 2, 2017 at 15:30

1 Answer 1

WCR = WeylCharacterRing(['A',3], style='coroots')
WR = WeightRing(WCR)
PR = WR.positive_roots()
q = prod([WR(1/2*x) - WR((-1/2*x)) for x in PR])
qs = q*q

105*A3(0,0,0) - 6*A3(0,2,0) - 15*A3(0,0,4) + 27*A3(0,1,2) - 45*A3(1,0,1) - 12*A3(1,2,1) + 6*A3(1,1,3) + 9*A3(0,4,0) - 3*A3(0,3,2) + 27*A3(2,1,0) - 9*A3(2,0,2) - 15*A3(4,0,0) - 3*A3(2,3,0) + A3(2,2,2) + 6*A3(3,1,1) - 3*A3(3,0,3)


  • $\begingroup$ Nice, I didn't know about WeylCharacterRing. However, since I want to compute $q^2$ in the $E_8$ case, this brute force approach can't work (there are $696\,729\,600$ terms in $q$). I'll clarify the question. $\endgroup$
    – Gro-Tsen
    Jun 1, 2017 at 21:18
  • $\begingroup$ @Gro-Tsen I see. My code is works perhaps for classical up to rank 4 or 5 and $G_2$. See the updated CoCalc worksheet. By the way, it seems that the resulting polynomial is invariant with respect to some Dynkin diagram automorphisms. $\endgroup$ Jun 1, 2017 at 22:02
  • 1
    $\begingroup$ @VítTuček, of course these symmetries are no accident; the square is $(-1)^{\lvert\Phi\rvert/2}\prod_{\alpha \in \Phi} (\exp(\alpha) - 1)$, which is manifestly symmetric under all automorphisms of $\Phi$. $\endgroup$
    – LSpice
    Jun 1, 2017 at 22:35
  • $\begingroup$ @LSpice Of course. I made an off-by-one error when I was looking for the triality symmetry at D3. :) $\endgroup$ Jun 1, 2017 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.